Asked by rei
An ore car of mass 41000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 9.6 m lower vertically, is a horizontally situated spring with constant 4.7 × 10^
5 N/m.
The acceleration of gravity is 9.8 m/s^
2. Ignore friction.
How much is the spring compressed in stopping the ore car? Answer in units of m.
5 N/m.
The acceleration of gravity is 9.8 m/s^
2. Ignore friction.
How much is the spring compressed in stopping the ore car? Answer in units of m.
Answers
Answered by
drwls
Gravitational potential energy loss = Spring potential energy increase
M g H = (1/2) k X^2
k = 4.7*10^5 N/m
M = 4.1*10^4 kg
H = 9.6 m
You known what g is.
Solve for X
M g H = (1/2) k X^2
k = 4.7*10^5 N/m
M = 4.1*10^4 kg
H = 9.6 m
You known what g is.
Solve for X
Answered by
rei
Thank you!
Answered by
rei
I got it wrong. I did exactly what you said. Multiplied m times g times h then multiplied by 2. Sq root that number and got 2778.926411 but it was wrong.
Answered by
drwls
X = sqrt(2 M g H/k)
= 4.54 m
You forgot to divide by k, the spring constant.
= 4.54 m
You forgot to divide by k, the spring constant.